Infer and justify non-trivial sequence patterns

Determine the underlying rule and give the missing terms; explain each in one or two sentences: a) 2, 3, 5, 9, 17, 33, ?; b) 1, 4, 2, 8, 4, 16, 8, 32, ?, ?; c) 1, 3, 7, 15, 31, ?; d) Complete the ? in the 3×3 grid where every row obeys the same rule: [[7, 4, 11], [5, 9, 14], [?, 13, 18]].

Overview: The problem tests mathematical reasoning, sequence inference, and pattern recognition by requiring the solver to find the governing rules and justify the missing terms in several numeric sequences and a small grid.

Solution General strategy: Look for (1) constant or patterned differences, (2) multiplicative/alternating operations, (3) powers/exponential forms, and (4) simple row-wise arithmetic relationships. Prefer the simplest rule that fits all given terms.

• (a) Rule: First differences double: $3-2=1$, $5-3=2$, $9-5=4$, $17-9=8$, $33-17=16$. Next difference is 32, so next term = $33 + 32 = 65$. Justification: The differences follow 1, 2, 4, 8, 16 (powers of two); equivalently $a_n = 2a_{n-1} - 1$.
• (b) Rule: Alternating operations: ×4 then ÷2 (1 → 4 → 2 → 8 → 4 → 16 → 8 → 32 →...). Next two: $32 \div 2 = 16$, then $16 \times 4 = 64$. Justification: The pattern strictly alternates between multiplying by 4 and halving.
• (c) Rule: Each term is previous ×2 + 1; equivalently the sequence is $2^n - 1$ ($n = 1, 2, 3, \ldots$). Next: $31 \times 2 + 1 = 63$. Justification: 1, 3, 7, 15, 31 match $2^1 - 1$, $2^2 - 1$, $2^3 - 1$, $2^4 - 1$, $2^5 - 1$.
• (d) Rule (row-wise): Third entry = first + second. Check: $7 + 4 = 11$; $5 + 9 = 14$; so $? + 13 = 18 \Rightarrow ? = 5$. Justification: The same row rule holds across all rows, yielding the missing value 5.

Validation and notes:

• Each proposed rule is consistent with all provided terms and yields a unique next value. While many rules can fit short sequences, the simplest consistent pattern is typically intended in interview settings.

Community answers Answer by yogyaburra a) 65 ($2^k + 1$, starting from $k = 0$) b) 16, 64 (odd terms = $2^k$, even terms = $4 \cdot 2^k$) c) 63 ($2^k - 1$, starting from $k = 1$) d) 5 ($a + b = c$ in $(a, b, c)$)